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JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - JEE MCQ


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30 Questions MCQ Test Mock Tests for JEE Main and Advanced 2025 - JEE Main 2020 January 8 Shift 1 Question Paper & Solutions

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation. The JEE Main 2020 January 8 Shift 1 Question Paper & Solutions questions and answers have been prepared according to the JEE exam syllabus.The JEE Main 2020 January 8 Shift 1 Question Paper & Solutions MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions below.
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JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 1

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB = (TA - 1.5) eV. If the de Broglie wavelength of these photoelectrons λB = 2λA, then the work function of metal B is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 1

de Broglie wavelength (λ),

On solving, TA = 2 eV
∴ KB = TA – 1.5 = 0.5 eV
∴ Work function of metal B is:
ϕB = EB – KB
= 4.5 – 0.5 = 4 eV
For A, ϕA  = EA – TA = 2 eV
So, work function for B is 4 eV

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 2

The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of 20 Ω, the null point on it is found to be at 1000 cm. The resistance of whole wire is:

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 2

5 = λl
Where λ is potential gradient & L is total length of wire

= 60 mA × R
R = 100 Ω

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JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 3

Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1 V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be:

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 3

In parallel combination, C1 + C2 = 10 μF
When connected across 1 V battery, then

∴ C2 = 8 μF, C1 = 2 μF
∴ In series combination,

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 4

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 4


At elongated position (x),
Fradial = mr ω2
∴ kx = m(ℓ + x)ω2

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 5

Boolean relation at the output stage-Y for the following circuit is:

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 5

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 6

Consider a solid sphere of radius R and mass density  The minimum density of a liquid in which it will float is:

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 6

For minimum density of liquid, solid sphere has to float (completely immersed) in the liquid.
∴ mg = FB (Also, Vimmersed = Vtotal)

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 7

Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m/s2 by an applied magnetic field (west to east). The value of magnetic field is:
(Rest mass of proton is 1.6 × 10-27 kg)

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 7


JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 8

The dimension of stopping potential V0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is:

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 9

A thermodynamic cycle xyzx is shown on a V-T diagram.

The P-V diagram that best describes this cycle is: (Diagrams are schematic and not to scale.)

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 9


From the corresponding V-T graph,
Process xy → lsobaric expansion,
Process xy → lsochoric (Pressure decreases)
Process z – x → Isothermal compression
Therefore, the corresponding PV graph is:

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 10

Consider two solid spheres of radii R1 = 1 m, R2 = 2 m and masses M1 and M2, respectively. The gravitational field due to sphere (1) and (2) is shown. The value of M1/M2 is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 10

From the diagram:
Gravitation field at the surface,

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 11

The plot that depicts the behaviour of the mean free time τ (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graphs are schematic and not drawn to scale.)

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 11

Relaxation time 

The graph of  is a straight line.

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 12

The graph which depicts the results of Rutherford gold foil experiment with α-particle is:
θ: Scattering angle
Y: Number of scattered α-particles detected
(Plots are schematic and not to scale.)

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 12

In 1911, Ernest Rutherford published a formula, which indicates that the number of particles (Y) that would be deflected by an angle 'θ' due to scattering is:

∴ Corresponding graph:

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 13

Three charged particles A, B and C with charges -4q, 2q and -2q are present on the circumference of a circle of radius d. The charged particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x-direction is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 13

The electric field due to B will be in the direction of line joining B and O (or B and C)

The electric field due to C will be in the direction of line joining O and C.
Net electric field due to B and C along
Now, electric field due to A along OA = 
Net electric field along the x-axis is 

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 14

In finding the electric field using Gauss law, the formula   is applicable. In the formula, ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situations?

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 14

According to Gauss's Law,

Where  is the electric field and  is the elemental.
areal vector for any small elemental area on the gaussian surface.

Given that the formula 

This relation will be valid when  is a constant and  and also when θ = 0°

The condition θ = 0 means that the components of electric field parallel to the elemental surface ara E1 sin θ = 0.
This means that the potential gradient over the surface is zero. Hence the surface is equipotential.

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 15

The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass 4 kg (The coordinates of the same are shown in figure) are

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 15


For the given lamina,
A1 = 1, C1 = (1.5, 2.5)
A2 = 3, C2 = (0.5, 1.5)

∴ Coordinates of centre of mass: (0.75, 1.75)

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 16

At time t = 0, magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 16

Using Faraday law,


Area = (16 × 4 - 2 × Area of triangle) cm2 

= 56 × 10-4 m2

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 17

The critical angle of medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 4/3 for this wavelength, will be

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 17

For relative permittivity
For relative permeability

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 18

The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye-piece?

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 18


For telescope,
Tube length (L) = fo + fes and magnification (m) = fe/fo , where fo and fe are focal lengths of objective and eye-piece
∴ fo + fe = 60 and fe = 5fo
∴ fo = 50 cm
fe = 10 cm

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 19

Consider a uniform rod of mass M = 4 m and length l pivoted about its centre. A mass m moving with velocity v making angle θ = π/4 to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 19


About point P, angular momentum (L) of the system,

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 20

A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion, is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C relative to the density at T = 0°C is close to? (Answer upto 2 decimal places)

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 20

Law of floatation:

In the given case,

*Answer can only contain numeric values
JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 21

Four resistances of 15 Ω, 12 Ω, 4 Ω and 10 Ω respectively in cyclic order to form Wheatstone's network. The resistance that is to be connected in parallel with the resistance of 10 Ω to balance the network is _________ Ω.


Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 21


Wheatstone bridge balance condition:

*Answer can only contain numeric values
JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 22

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP in 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________ Hz.


Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 22

vair = 300 m/s

*Answer can only contain numeric values
JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 23

A body A of mass m = 0.1 kg has an initial velocity of . It collides elastically with another body B of the same mass which has an initial velocity of . After collision, A moves with a velocity  The energy of B after collision is written as x/10 J. The value of x is ______.(Nearest Integer)


Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 23

By conservation of momentum,

*Answer can only contain numeric values
JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 24

A particle is moving along the x-axis with its coordinate with time 't' given by x(t) = 10 + 8t - 3t2. Another particle is moving along the y-axis with its coordinate as a function of time given by y(t) = 5 - 8t3. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as √v. Then v (in m/s) is _______.(Nearest Integer)


Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 24


At t = 1 s,

Therefore,
v = 580 (m/s)

*Answer can only contain numeric values
JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 25

A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of the lens material is 1.5, then the focal length of the lens (in cm) is _______.


Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 25

Lens maker formula:

For plano-convex lens,

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 26

The complex that can show fac-and mer-isomers is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 26

Octahedral complexes with the general formula of [MX3Y3] can show fac-mer isomerism.
[Co(NH3)3(NO2)3] will show fac-and mer-isomers.

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 27

As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 27

Fe(OH)3 is a positive sol. Its coagulation will be caused by the anion of the electrolyte.
The flocculation value is inversely proportional to the coagulation power or valency of the anion.
The correct order of flocculation values is:
K3 [Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 28

The decreasing order of reactivity towards dehydrohalogenation (E1) reaction of the following compounds is

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 28

Dehydrohalogenation of the given halides by E1 mechanism is decided by the stability of the carbocation formed in the first step. The correct decreasing order of the given halides towards dehydrohalogenation by E1 mechanism is:

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 29

For the Balmer series in the spectrum of H atom,  the correct statements among (l) to (IV) are:
(I) As wavelength decreases, the lines in the series converge.
(II) The integer n1 is equal to 2.
(III) The lines of longest wavelength correspond to n2 = 3.
(IV) The ionisation energy of hydrogen can be calculated from wave number of these lines.

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 29

In the Balmer series of H-atom, electronic transitions take place from higher orbits to 2nd orbit, and the longest wavelength will correspond to the transition from 3rd orbit to 2nd orbit.
∴ n1 = 2 and n2 = 3 for longest wavelength.
As wavelength decreases, the lines in the Balmer series converge. The correct statements are (I), (II) and (III).
The ionisation energy of hydrogen can be calculated from wave number of the last line of the Lyman series:
n1 = 1 and n2 = ∞
Hence, statement IV is incorrect.

JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 30

The first ionisation energy (in kJ/mol) of Na, Mg, Al and Si, respectively, is:

Detailed Solution for JEE Main 2020 January 8 Shift 1 Question Paper & Solutions - Question 30

Ionisation energy of the elements belonging to period III in general increases as we move from left to right with the exception of group-2 and group-15 elements due to their stable configuration. The increasing order of the first ionisation energy of the given elements is:
Na < Al < Mg < Si.
Ionisation energy of the given metals:
Na: 496 kJ/mol; Al: 577 kJ/mol
Mg: 737 kJ/mol; Si: 786 kJ/mol

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